A student applies a potential difference V of (4.0 ± 0.1) V across a resistor of resistance R of (10.0 ± 0.3) Ω for a time t of (50 ± 1) s.

Question 20

A student applies a potential difference V of (4.0 ± 0.1) V across a resistor of resistance R of (10.0 ± 0.3) Ω for a time t of (50 ± 1) s.

The student calculates the energy E dissipated using the equation below.

E = V²t / R = (4.0² × 50) / 10.0 = 80 J

What is the absolute uncertainty in the calculated energy value?

A 1.5 J                        B 3 J                           C 6 J                           D 8 J

Reference: Past Exam Paper – June 2018 Paper 12 Q4

Solution:

Answer: D.

E = V²t / R

ΔE / E = 2(ΔV/V) + Δt/t + ΔR/R

ΔE = [2(ΔV/V) + Δt/t + ΔR/R] × E

ΔE = [2(0.1/4.0) + (1/50) + (0.3/10.0)] × 80

ΔE = 0.1 × 80

ΔE = 8 J